# CRN 11378: HW 5

From Classes

(Difference between revisions)

HelmutKnaust (Talk | contribs) |
HelmutKnaust (Talk | contribs) |
||

Line 12: | Line 12: | ||

'''Problem 24.''' A function $f:\mathbb{R} \to \mathbb{R}$ is called ''bounded'' if there is an $M>0$ such that $|f(x)|\leq M$ for all $x\in\mathbb{R}$. | '''Problem 24.''' A function $f:\mathbb{R} \to \mathbb{R}$ is called ''bounded'' if there is an $M>0$ such that $|f(x)|\leq M$ for all $x\in\mathbb{R}$. | ||

#Let $f,g:\mathbb{R}\to\mathbb{R}$ be two bounded functions that are uniformly continuous on $\mathbb{R}$. Show that $f\cdot g$ is uniformly continuous on $\mathbb{R}$. | #Let $f,g:\mathbb{R}\to\mathbb{R}$ be two bounded functions that are uniformly continuous on $\mathbb{R}$. Show that $f\cdot g$ is uniformly continuous on $\mathbb{R}$. | ||

− | #Show that the result | + | #Show that the result may fail without the boundedness condition. |

## Revision as of 11:29, 5 November 2019

**Problem 21.** Let the function $f:\mathbb{R}\to\mathbb{R}$ be given by $f(x)=\sqrt[3]{x}$.

- Show that $f$ is continuous at $0$.
- Show that $f$ is continuous at any $x_0\neq 0$. (The identity $a^3-b^3=(a-b)(a^2+ab+b^2)$ will be helpful.)

**Problem 22.** Assume $f:\mathbb{R}\to\mathbb{R}$ is continuous on $\mathbb{R}$. Show that $\{x\in\mathbb{R}\ |\ f(x)=0\}$ is a closed set.

**Problem 23.** Let $c\geq 0$. Assume $f:\mathbb{R}\to\mathbb{R}$ satisfies $|f(x)-f(y)|\leq c\cdot |x-y|$ for all $x,y\in\mathbb{R}$.

- Show that $f$ is uniformly continuous on $\mathbb{R}$.
- Now assume that $0<c<1$. Show that there is an $x\in\mathbb{R}$ such that $f(x)=x$. (Hint: for any $y\in\mathbb{R}$ look at the sequence $y,f(y),f(f(y)),f(f(f(y)))\ldots$.
- Show that the result in 2. above may fail if $c=1$.

**Problem 24.** A function $f:\mathbb{R} \to \mathbb{R}$ is called *bounded* if there is an $M>0$ such that $|f(x)|\leq M$ for all $x\in\mathbb{R}$.

- Let $f,g:\mathbb{R}\to\mathbb{R}$ be two bounded functions that are uniformly continuous on $\mathbb{R}$. Show that $f\cdot g$ is uniformly continuous on $\mathbb{R}$.
- Show that the result may fail without the boundedness condition.